Brachistochrone with velocity - still a cycloid

[Grand]

Homework Statement

Brachistochrone problem: if the particle is given an initial velocity v_0 \neq 0[\tex] , show that the path of minimum time is still a cycloid.<br /> <br /> <h2>Homework Equations</h2><br /> Conservation of energy:<br /> \frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2[\tex]&lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; I know how to start the problem, but in the end have to solve the differential equation:&lt;br /&gt; \frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}[\tex]&amp;lt;br /&amp;gt; which I can&amp;amp;#039;t solve. Any ideas and hints would be greatly appreciated!

 

[Grand]

Anyone with a decent solution?

 

[Mandeep Deka]

Check this out, see if it helps!

http://mathworld.wolfram.com/BrachistochroneProblem.html

 

[Andrew Mason]

Homework Statement

Brachistochrone problem: if the particle is given an initial velocity v_0 \neq 0 , show that the path of minimum time is still a cycloid.

Homework Equations

Conservation of energy:
\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2

The Attempt at a Solution

I know how to start the problem, but in the end have to solve the differential equation:
\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}
which I can't solve. Any ideas and hints would be greatly appreciated!

You have to put at forward slash to close a latex instruction ie: /tex

Think of the initial velocity as the result of a vertical change in position of a height y0 along this path from a starting point of 0. Then you just have:

\frac{dy}{dx}=\sqrt{\frac{k^2-2gy0-2gy}{2gy0+2gy}}

which reduces to:

\frac{dy}{dx}=\sqrt{\frac{k^2-2g(y0+y)}{2g(y0+y)}}

\frac{dy}{dx}=\sqrt{\frac{k^2-2gy&#039;}{2gy&#039;}}

the solution to which is a cycloid.

AM